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On  the  Cardioids 

FULFILLING 

Certain  Assigned  Conditions 


By 

SisTEH  Mary  Gervase,  m.a. 

of 

THE    SISTERS   OF   CHAHITY,    H  4HPAX,    N.    S. 


A  DISSERTATION 

Submitted  to  the  Catholic  Sisters  College  of  the  Catholic 

University  of  America  in  Partial  Fulfillment 

of  the  Requirements  for  the  Degree 

Doctor  of  Philosophy 


Washington,  D.  C. 
June,  1917 


"if 


On  the  Cardioids 

FULFILLING 

Certain  Assigned  Conditions 


Sister  Mary  Gervase,  M.A. 

of 

THE    SISTERS    OF    CHARITY,    HALIFAX,    N.    S. 


A  DISSERTATION 

Submitted  to  the  Catholic  Sisters  College  of  the  Catholic 

University  of  A  merica  in  Partial  Fulfillment 

of  the  Requirements  for  the  Degree 

Doctor  of  Philosophy 


Washington,  D.  C. 
June,  1917 


NATIONAL  CAPITAL  PRESS,  INC.,  WASHINGTON,   0.  C. 


Contents 


PAGE 

Introductory 5 

The  Simpler  Cases: 

I.  Given  3  Conditions 9 

II.  Given  4  Conditions 15 

More  Difficult  Problems: 

I.  Given  3  Conditions 24 

II.  Given  4  Conditions 35 


362099 


Introductory 


Tlie   triciispidal,  bicircular  qiiartic  of  the  third  class  defined 

hy  the 

Cartesian  equation  (.i'-+?/-+a.r)'"  =  a-(.r'^+?/") 
polar  equation  p  =  a  ( 1  —  cos  6) 

and  commonly  known  as  the  Cardioid,  has  for  many  years  been 
the  object  of  mathematical  investigation.  It  has  lately  been 
studied  by  Raymond  Clare  Archibald  in  his  Inaugtiral-Dissertation 
"The  Cardioide  and  Some  of  Its  Related  Curves"  (Strassburg, 
1900),  which  work  contains  an  historical  sketch  of  the  curve  and 
a  presentation  of  results  prior  to  the  year  of  its  publication. 
Since  then,  the  only  work  on  the  subject  of  considerable  length  is 
Professor  Archibald's  paper,  "The  Cardioid  and  Tricuspid :  Quar- 
tics  with  Three  Cusps."*  Besides  this  there  have  appeared  a  few 
detached  problemsj  and  contributions  in  periodicals  treating  the 
curve  from  either  a  metric  or  a  projective  standpoint. 

The  chief  characteristic  of  former  research  along  this  line  seems 
to  be  the  examination  of  the  cardioid  as  a  fixed  curve  and  the 
consideration  of  its  properties  as  such.  The  present  investigation 
starts  from  a  different  point  of  view,  which  we  may  outline  as 
follows : 

In  general,  a  curve  of  the  fourth  degree  is  capable  of  satisfying 
14  conditions.  The  cardioid,  however,  having  3  cusps,  two  of 
wjiich  are  at  the  fixed  (circular)  points  /  and  J,  can  be  subjected 
to  only  4  conditions.!  If,  then,  3  conditions  be  imposed,  there 
are  oc  ^  curves  satisfying  them;  therefor^,  the  special  elements 
(cusp,  focus,  double  tangent)  describe  definite  loci.  If  4  conditions 
are  given,  there  are  a  finite  number  of  curves  satisfying  them.  It 
is  our  purpose  to  obtain  the  loci  generated  in  the  first  case;  and, 
in  the  second,  to  determine  the  number  and  (where  possible)  the 
reality  of  cardioids  for  various  kinds  of  assigned  conditions. 

The  co-ordinate  system  which  lends  itself  most  readily  to  such 
an  investigation  is  the  system  of  conjugate  (also  called  circular) 
co-ordinates,  in  which  a  point  is  named  by  a  vector  which  has  its 


*  AnnaU-  of  Mathematics:  M  S.  4.  190^2-3,  pp.  9off. 

t  Cf.,  for  example,  Questions  14,4^)5;    1().^2G();    10.^}9"2.     Ed.  Times  (London ) . 
i  Cusps  at  I  and  J  count  for  4  conditions  each.      The  third  cusp  counts  for 
2  more  conditions.     Thus,  the  14  conditions  reduce  to  14  — (2X4  +  2)  -=4. 


6  CanUoids  FulfUliiKj  Crifdin  Assigned  CoiuUtions 

initial  extremity  at  some  fixed  point,  called  the  origin.  Denoting 
a  vector  by  z,  its  projections  on  the  real  and  the  imaginary  axes 
are  designated  by  x,  ?/,  respectively;  and 

z  =  x-\-iy.'^ 

With  a  vector  z  is  associated  its  conjugate  z' .  This  may  be  geo- 
metrically defined  as  its  reflection  in  the  real  axis.     Accordingly, 

z'  =  x  —  iy. 

All  vectors  may  be  considered  as  obtained  from  the  standard 
unit  vector  by  means  of  a  stretch  and  a  turn.  Turns  are  regularly 
designated  by  the  letter  /;  fixed  turns,  by  /i,  ^2,  ^3  •  •  • ;  variable 
turns,  by  ^,  ^^  r   .    .    .  f 

Regarding  the  cardioid  as  the  epicycloid  generated  by  equal 
circles,  the  map-equation  of  the  curve  is 

The  centre  of  the  fixed  circle  is  at  0,  which  point,  we  shall,  with 
Professor  Morley,|  call  the  centre  of  the  cardioid.  This  point  is 
also  the  singular  focus  of  the  curve.  The  cusp  is  at  z=^r  (r  being 
on  the  real  axis). 

This  map-equation  involves  its  conjugate. 


'"('-;.)■ 


These    two    equations    taken    together    may    be    considered    the 
parametric  equations  of  the  curve. 

A  cardioid  with  centre  at  Zo  and  any  orientation  has  for  its 
map-equation : 

z  —  Zo  =  2at  —  a'f 

z  =  Zo-]-—  gives  the  cusp. 

The  angle  made  by  the  axis  of  the  curve  with  the  axis  of  reals  is  0, 
where 


,^ie 


{:)■■ 


*  The  symbol  i  denotes,  as  usual,  V  —  1. 

f  Cf.  Morley,  "On  Reflexive  Geometry,"  Transcicfions  of  the  American 
Mdfhematical  Society,  1907,  Vol.  8,  p.  14. 

J  "Metric  Geometry  of  the  Plane  X-Liiic""  T ran.sncllonn  of  the  American 
Mathematical  Society,  1900,  Vol.  1,  p.  10.5. 


Cardioids  I'ldfiWuif/  Certain  Assigned  Conditions  7 

Any  tangent  is  given  by 

{z-Zo)-{z'-z'o)t'-3at+Sa't^  =  0; 
the  double  tangent,  by 

a''{z-Zn)+a'(z'-z'o)=Sa'a". 

The  kinds  of  data  we  sliall  consider  are:  point,  line,  centre,  cusp, 
or  double  tangent  given,  to  solve  a  specified  problem.  The  last 
three  are  equivalent  to  2  conditions  each.  The  problems  arising 
naturally  fall  into  2  classes: 

I.  Given  3  conditions,  find  specified  loci;  as, 

(1)  Given  the  centre  and  a  line;  find  the  locus  of  cusps. 

(2)  Given  the  centre  and  a  ])oint;  find  the  locus  of  cusps. 
(8)  Given  the  cusp  and  a  line;  find  the  locus  of  centres. 

(4)  (iiven  the  cusp  and  a  ])oint;  find  the  locus  of  centres. 

(5)  Given  the  double  tangent  and  a  line;  find  the  locus  of  cusps. 
(Jiven  the  double  tangent  and  a  line;  find  the  locus  of  centres. 

(0)  (iivcMi  the  double  tangent  and  a  point;  find  the  locus  of  cusps. 
Given  the  double  tangent  and  a  i)()int;  find  the  locus  of  centres. 

(7)  Given  8  hues,  find  the  locus  of  centres. 

(8)  Given  "i  lines  and  1  point,  find  the  locus  of  centres. 

(9)  Given  1  line  and  2  points,  find  the  locus  of  centres. 

(10)  (liven  3  points,  find  the  locus  of  cus})s. 

II.    Given    4  conditions,    find    how    many    solutions    there 

are;  as, 

(a)  (liven  the  centre  and  2  lines,  how  many  cardioids  are  there.^ 

TIow  many  are  real? 
(}))  (liven  the  centre,  a  point  and  a  line.     Apj)ly  the  same  ques- 
tions. 

(c)  Given  the  centre  and  2  ])oints. 

(d)  (iiven  the  cusp  and  2  lines. 

(e)  (jiven  the  (tusj),  1  line  and  1  })oint. 

(f)  Given  the  cusp  and  2  ])()ints. 

(g)  Given  the  double  tangent  and  2  lines. 

(h)  (liven  the  double  tangent,  1  line  jind  1  point. 

(i)  Given  the  dou})le  tangent  and  2  i)oints. 

(j)  (liven  4  lines. 

(k)  Given  .'J  lines  and  I  point. 

(1)  Given  2  lines  and  2  j)()ints. 


8  Vaidiohh  Fulfilling  Certain  Assigned  Conditions 

(ill)  Given  1  line  and  3  points. 
(n)  Given  4  points. 

Auain,  tliere  is  an  evident  division  of  the  problems  into  simpler 
and  more  difficult  cases.  It  is  not  the  intrinsic  value  of  some 
of  the  simple  cases  which  authorizes  their  appearance  in  this 
work,  but  rather  their  importance  in  the  solution  of  some  of  the 
more  difficult  problems. 

Before  taking  up  the  problems,  it  will  be  well  to  indicate  a 
guiding  ])rinciple  which  will  be  found  of  great  importance  for 
cases  where  the  cardioids  are  to  touch  several  lines.  It  may  be 
stated  thus: 

If  d  be  the  angle  between  2  lines  which  touch  a  cardioid,  the 
angle  between  the  cuspidal  rays  to  the  points  of  tangency  has 

some  one  of  the  values  %,  f  ((9+360°),  ^((9+7^20"). 

For  the  case  of  parallel  tangents,  this  specializes  to  the  well- 
known  theorem: 

The  points  of  contact  of  any  8  i:)arallel  tangents  lo  a  cardioid 
subtend  angles  of  120°  at  the  cusp. 

With  these  considerations  premised,  we  shall  proceed  to  a 
treatment  of  the  simpler  cases. 


The  Simpler  Cases 


I.  Given  3  Conditions. 
PROBLEM  (1):  Given  the  center  and  a  line;  find  the  locus  of 
cusps. 

Let  us  take  the  centre  at  0;  the  line,  as  2+2' =  2. 
The  map-equation  of  the  cardioid  is  z  =  2at—a'f 

The  cusp:  z  —  —r 
a 

Any  tangent:  z-Sat-i-SaT-z't^  =  0 

If  the  line  2+2'  =  2  is  to  touch  the  cardioid,  the  following  iden- 
tity must  exist: 

z-Sat+Sa'f-z't^^  z+z'-2: 

whence,  ^^  =  —  1 (1) 

and  -Sat-\-Sa'f=  -2 (2) 

From  (2)  we  get  t=  —1,  -co,  or  — oj^* 

It  will  be  sufficient  to  take  one  of  these  values;  for,  since  the 

expression  for  the  cusp  I  -,  I  does  not  occur  rationally  in  (2),  the 


rationalized  form  of  the  equation  will  be  the  same,  no  matter 
which  value  is  selected  for  t.  Cubing  (2)  and  substituting  the 
value  ^=  —  1,  we  obtain 

27(a3+a'3)  -54aa'+8  =  0. 
Now  the  cusp  is  given  by  2 =-7,  and  we  seek  the  cusp  locus. 
Therefore,  if  we  call  — -„  k  [whence  result  the  equations: 

a 

aa'  =  kk' 
a^^a'^  =  kk'{k+k')l 

the  equation  of  the  locus  becomes 

nkk'{k+k')-54^kk'-]-S  =  0. 

This  is  a  rational  cubic  with  its  double  point  at  A;=-  and  vertex 


*co  and  co^  are  the  imaginary  cube  roots  of  unity. 


10         Cardioids  FulfiUing  Certain  Assi(/nr(J  CotiditioH, 


at  A:  =  --   (Fig.  I).  The  line  2+2'  =  2  is  an  asymptote  to  the  curve. 

Making  the  equation  homogeneous,  it  becomes 
27kk\k-\-k'-2iv)  +8w^  =  0, 
which  is  of  the  form 

In  this  form  it  is  easily  seen  that  the  curve  has  points  of  inflection 
at  the  intersections  of  A:  =  0  and  w  =  0,  and  of  A''  =  0  and  w?  =  0. 


Fig.  I. 

respectively;  i.  e.,  at  the  circular  points,  /  and  J.      Moreover, 
A;  =  0  is  the  flex  tangent  at  /;  k^  =  0,  the  flex  tangent  at  J. 
If  the  given  line  be  taken  to  be 

z  =  z'ti-\-au 
the  cusp  locus  becomes  the  circular  cubic 

^7tikk\k-k'ti-ai)-ai^  =  0, 
which  also  has  points  of  inflection  at  /  and  J. 

PROBLEM  (2):  Given  the  centre  and  a  point;  find  the  locus  of 
cusps. 

Let  us  take  the  centre  at  0;  the  point,  as  z=  L 
Map-equation  of  the  cardioid:  z  =  ^at  —  a'f. 
Since  the  cardioid  is  to  be  on  the  point  z=  1,  we  have  the  equation 


*  Cf.   Salmon,   Analytische   Geomeirie  der  hoheren  ebenen   Kurven,  Leipzig, 
1882  (Zweite  Auflage).  p.  50. 


Cardioids  Fulfilling  Certain  Assigned  Conditions        11 

which  involves  its  conjugate 

•t      ^2-1- 
From  these  two  equations  it  is  necessary  to  eliminate  t  and  express 
the  result  in  terms  of  -7,  which,  as  before,  we  shall  designate  as  k. 
The  elimination  of  t  gives 

3aV2+6aa'-4(a3+a'3)  =  1. 

In  terms  of  k,  k'  the  locus  is 

Sk%'^+6k¥-^kk\k-{-k')  =  1. 

This  bicircular  quartic  (see  Fig,  IV,  p.  20)    has  its  vertex  at  — - 

and  Cusp  at  1.     Expressed  in  homogeneous  form 

kk\Skk'-4>kw-Ww-\-6w^)-iv'  =  0. 

it  is  easy  to  see  that  the  curve  intersects  k  =  0,  k'  =  0  in  4  coincident 
points.     The  form 

¥%{Sk-^w)-^w{Qk¥w-^k''k'-w'')  =  0 

shows  that  the  curve  goes  through  /,  the  tangents  thereat  being 
A*  =  0  and  Sk  —  'iw  =  0.  But  since  there  are  4  coincident  points  at  /, 
k  =  (i  must  be  a  flex  tangent;  that  is,  /  is  a  flecnode.  Similarly, 
the  tangents  at  J  are  A:'  =  0  and  3A*'  — 4w'  =  0,  the  former  being 
a  flex  tangent. 

When  J)  is  the  given  point,  the  cusp  locus  is  the  bicircular  quartic 

S¥k'^-^kk\kp'^k'j))-\-Qpp'kk'-p^p'''  =  (), 

which  has  flecnodal  points  at  /  and  J  with  A;  =  0  and  A:'  =  0  as  flex 
tangents. 

PROBLEM  (3):  Given  the  cusp  and  a  line;  find  the  locus  of 
centres. 

Let  the  cusp  be  at  0;  the  line,  2+2'  =  2. 

The  equation  of  a  cardioid  with  cusp  at  6  is 

z-\-^,  =  ^at-a't^\ 
a 

{a-a'ty 
I.e.,  z=-- r-^ 


a 


The  centre  is  given  by  2=  — , 


12         Cardioids  Fulfilling  Certain  Assigned  Conditions 
Any  tangent  to  this  is 


z-z'P-Sat+3a't^-^-^+   ,  =  0. 
a       a 


If  2+2^  =  2  is  to  be  tangent, 


'2^3  2 

z-z't'-Sat+Sa'f-^-^+-,  =z-\-z'-2. 
a        a 


From  this  identity,  it  follows  that 
and 


(1) 
(2) 


From  (1),  /=  — 1,  —CO,  or  — oj^ 

Cubing  (2)  and  substituting  the  value  t=  —1,  we  obtain 

27[a3+a'3+3aa'(a+aO]=  -^  -—-2. 
Calling },  c; ,  c',  the  equation  of  the  locus  reduces  to 


(c+c'-2)3+54cc'  =  0. 


Fig.  II. 
This  is  a  nodal  cubic,   known  as  the  Tschirnhausen  Cubic,* 

with  vertex  at  -  and  double  point  at  —2  (Fig.  II). 
4 

If  the  given  line  is  z  =  z^ti-\-aiy  the  centre  locus  becomes  the 

nodal  cubic  (c  —  c'ti  —  ai)^  —  27cc'aiti  =  0. 

PROBLEM  (4):  Given  the  cusp  and  a  point;  find  the  locus  of 
centres. 

Let  the  cusp  be  at  0;  the  point,  at  z=l. 


*  It  is  also  called  the  Cubique  de  I'Hopital.  Cf.  Archibald,  op.  cit.,  p,  19, 
where  he  states  that  "the  locus  of  the  vertices  of  co-cuspidal  cardioides 
tangent  to  a  given  line  is  a  Tschirnhausen  Cubic." 


Cardioids  Fulfilling  Certain  Assigned  Conditions        13 

The  equation  of  a  cardioid  with  cusp  at  0  is 

(a-a'ty 


z=  —■ 


(centre :  c  = r\ 
a' ) 


a 
The  conditions  that  2;  =  1  be  on  the  cardioid  are  expressed  by 

2at-a'f--,  =1 

2^'-l-^  =1 
t       V-      a 

EHminating  t  from  these  equations  and  expressing  the  resulting 
equation  in  terms  of  c  and  c',  we  obtain  as  the  sought  centre  locus: 

4cc'=(c+c'-l)2 

This  is  a  parabola  with  vertex  at  -  and  focus  at  0. 

When  p  is  the  given  point,  the  locus  is  the  parabola  confocal 
with  the  preceding: 

^cc'pp'  =  (cp'-\-c'p—pp'y. 

PROBLEM  (5):  Given  the  double  tangent  and  a  line;  find  the 
locus  of  centres. 

Let  the  given  line  be  z  =  tiz';  the  double  tangent,   2+2'  =  2. 
The  equation  of  the  cardioid  may  be  taken  to  be 

z—Zo  =  r{2t  —  i'^),  where  r  is  real. 

That   the  given  conditions  be  fulfilled,  the  following  identities 
must  exist: 

(z-zo)  -  {z' -z'o)t^-Srt(l-t)  ^z-z'ti (1) 

(z-Zo)  +  (z' -z'o)  -Sr  ^  z-\-z' -2 (2) 

From  (1)   P  =  ti   [whence,  t  =  T,  cor,  coV,  where  r  =  ^V#i] 

and  Zo-^z'o-{-3r  =  2. 

Combining  these  conditions  and  eliminating  r  and  ty  we  find  the 

locus  breaks  up  into  the  3  lines  given  by  the  equations : 

Ml-r) 


Zo  =  Z  o  T 


Zo  =  z'oOJT-{ 


1-T  +  t2 

2coT(a)r— 1) 


Zo  =  Z^oCxi^T-\ 


l+coV^  — cor 
2cot{t  —  c*)) 


These  three  lines  pass  through  the  point  of  intersection  of  z  =  z^ti 


34         Cardioids  Fulfilling  Certain  Assigned  Conditions 

and  z+z'  =  2.     If  6  be  the  inclination  of  the  given  line,  z  =  z'tu 
the  first  line  of  the  centre  locus  is  inclined  to  the  axis  of  reals  at  an 

angle  equal  to  -;  the  others  at  angles  — - — ,  — - — ,  respectively. 

The  locus  of  cusps  of  cardioids  which  have  z-\-z'  =  2  as  double 
tangent  and  touch  the  line  z  —  z'ti  =  0  consists  of  3  lines  through 
the  intersection  of  2+2'=  2  and  2  — z'^i  =  0  such  that  if  a  be  the 
angle  the  centre  locus  makes  with  the  double  tangent,  and  /8  be 
the  angle  the  cusp  locus  makes  with  the  same  tangent,  then 


/3  =  tan 


(Jtanay 


PROBLEM  (6):  Given  the  double  tangent  and  a  point;  find  the 
locus  of  cusps. 

Let  us  take  the  point  at  the  origin,  and  the  equation  of  the 
cardioid  as: 

z  =  Zo-\-r  —  r-\-r{^t  —  t^)y  where  r  is  real, 

=  k-r(l-ty    ■ 

For  the  fulfillment  of  the  given  conditions,  the  following  identities 
must  exist: 

{z-k)-\-(z'-k')-r  z  2+2'-2 (1) 

k-r{l-ty  =  0 -...(2) 

k'-r{  l-]y  =  0 (3) 

From  (1)  k-\-k'+r  =  2 (4) 

Eliminating  t  and  r  from  (2),  (3),  (4),  we  obtain  as  the  required 
locus : 

4{^2-k-kykk'=[(-k-k'){^-k-k')+kkj 

As  an  aid  in  the  discussion  of  the  curve,  let  us  transform  it  to 
Cartesian  co-ordinates  by  means  of  the  equations 

k  =  x-{-iy] 
k'=^x—iyy 

The  equation  becomes 

9a:^-6a:V+2/^+24j-?/2-8a:3- 162/2  =  0 

This  is  of  the  form 

r/V+a:3(9x-8)  =  0, 

which  indicates  a  cusp  at  the  origin  with  2/  =  0  as  the  cusp  tangent; 
the  curve  has  also  a  branch  cutting  the  X-axis  at  the  point  I  q»  0  1. 


Cardioifh  FulfiUhif/  Certain  Assigned  Conditions         15 

Noting  the  behavior  of  the  curve  at  infinity,  we  observe  that  it 
has  2  paraboHc  branches  with  directions  determined  hy  y=  =tV3x; 
i.  e.,  the  infinite  branches  tend  towards  angles  of  ±120°  with  the 
X-axis. 

Where  p  is  the  given  point,  the  resulting  locus  takes  the  form: 

4{^-k-kyik'-p'){k-p)  =  [(2-k-k'){p'-k'+p-k) 

+{p'-k'){p-k)]\ 

which  has  one  cusp  at  p  and  2  parabolic  branches  with  directions 

.  k      1  ±V3i 

aetermmed  by  , /  =  — - — . 

The  locus  of  centres  of  cardioids  which  have  z-\-z'  =  2  for  double 
tangent  and  pass  through  the  origin  is  the  quartic: 

3x^-lSxY-\-27y'+  Wx^-^  lUxy^+24>x^-72y^-  16  =  0. 
This  curve  has  2  parabolic  branches  tending  towards  angles  of 
±150°  with  the  X-axis.     From  the  form 

2/2(27!/2-18x2+144a!-72)  +  (.r+2)3(3x-2)  =  0, 

it  is  easily  seen  that  the  curve  has  a  cusp  at  the  point  (  —  2,  0) 
with  y  =  0  as  the  cusp  tangent,  as  well  as  a  branch  cutting  the 

X-axis  at  the  point  I  q'  ^  )• 

The  foregoing  complete  the  simpler  cases  when  3  conditions  are 
given.     Let  us  now  proceed  to  the  cases  arising  from  them. 

II.  Given  4  Conditions. 

In  these  problems  we  shall  first  endeavor  to  find  the  number  of 
cardioids  which  fulfil  4  conditions  as  variously  assigned.  In  the 
cases  here  treated  this  number  is  found  by  the  intersections  of 
the  various  loci  already  obtained.  Such  a  solution  is,  however, 
the  maximum  number  and  includes  the  imaginary  curves,  if  there 
are  any.  It  will  therefore  be  our  next  problem  to  separate  these 
imaginary  solutions  and  thus  determine  the  number  of  real 
cardioids  for  each  case. 

PROBLEM  (a) :  Given  the  centre  and  2  lines. 

Let  the  given  centre  be  at  0;  the  2  lines,  z+z'  =  2  and  z  =  z'ti-\-ai. 
We  seek  the  number  of  cardioids  which  have  0  for  centre  and 
touch  each  of  the  two  given  lines. 

The  cardioids  which  have  0  for  centre  and  touch  the  line  z-\-z'  =  2 
have  their  cusps  on  the  circular  cubic 

27kk\k-\-k') -54kk' -\-S  =  0. 


16         Cardioids  Fulfilling  Certain  Assigned  Conditions 

Similarly,  the  cardioids  having  0  for  centre  and  touching  the  line 
z  =  z'ti-{-ai  have  their  cusps  on  the  circular  cubic 

27tikk'{k-k'ti-ai)-ai^  =  0. 

Thus,  the  cusps  of  cardioids  which  have  0  for  centre  and  touch 
both  of  the  given  lines  are  given  by  the  points  of  intersection 
of  these  two  circular  loci.  These  cubics  have  9  intersections. 
Therefore,  there  must  be  9  cusps  of  cardioids,  and  hence  (since 
cusp  and  centre  uniquely  determine  a  cardioid)  9  cardioids, 
which  fulfill  the  required  conditions.  But,  since  the"  loci  have 
points  of  inflection  at  the  circular  points  and  the  same  tangents 
thereat,  3  of  their  intersections  are  at  each  of  these  imaginary 
points.  Thus,  6  of  the  9  cardioids  are  imaginary.  The  other  3 
are  always  real.     For,  considering  the  two  given  lines,  either 

I.  They  will  be  equidistant  from  the  given  centre;  or, 

II.  One  will  lie  nearer  that  point  than  the  other.  Now  the 
circle  with  radius  equal  to  the  minimum  radius  vector  of  either 
cubic  lies  wholly  within  the  loop  of  that  cubic.  The  vertex  of  the 
second  cubic  will,  then,  lie  either  on  the  circumference  of  this 
circle  (Case  I)  or  within  it  (Case  II).  In  either  case,  this  point 
lies  within  the  loop  of  the  first  cubic.  The  second  curve,  in  tending 
towards  its  asymptote,  must  intersect  the  loop  of  the  first  in  2 
real  points,  and  only  in  2.  But  if  there  are  but  3  possible  real 
intersections,  and  two  have  been  shown  real,  the  third  must 
necessarily  be  real.  Thus,  there  are  3  real  cardioids  which  have 
a  given  centre  and  touch  2  given  lines.     (See  Fig.  X,  p.  44.) 

PROBLEM  (b) :  The  centre,  a  point  and  a  line  given. 

Take  the  origin  as  centre;  z-\-z'  =  2  as  the  given  line  and  />  as 
the  given  point. 

The  intersections  of  cusp-loci  will,  as  before,  give  the  number  of 
solutions.  The  cusp-locus  for  cardioids  with  centre  0  and  touching 
z-\-z'  =  2  is  the  circular  cubic 

27kk'(k-\-k')-5ikk'-\-S  =  0; 

the  cusp-locus  for  cardioids  with  centre  0  and  passing  through  p 
is  the  bicircular  quartic 

Sk''k'^-6pp'kk'-^kk\kp'-\-k'p)-py^  =  0. 

These  two  curves  have  12  common  intersections;  i.  e.,  there  should 
be  12  cardioids  satisfying  the  required  conditions.     But,  since 


Cardioids  Fulfilling  Certain  Assigned  Conditions        17 

the  two  curves  have  4  points  in  common  at  each  of  the  circular 
points,  there  can  be,  at  most,  4  real  cardioids.  However,  there 
are  not  always  4  real.  Therefore,  let  us  determine  the  regions  of 
the  plane  where  there  are  4,  or  only  2,  or  no  real  cardioids  satis- 
fying the  given  conditions. 

The  map  equation  of  any  cardioid  with  centre  at  0  is 

z  =  ^at-aT     , 

Since  2+2' =  2  is  to  be  tangent  to  this, 

z-Sat+Sa'f-z't^  -s+z'-^ 

Employing  the  value  t=  —1,  we  have  for  a,  a\  any  one  of  the  8 

equations : 

3a+3a/=-2 
3aco+3aV=-2 
3aco2+3a'a;=-2 

It  will  be  sufficient  to  take  the  first  equation.  Since  p  is  the 
given  point,  we  have 

^at-a't^  =  p 

whence,  eliminating  a,  a',  we  have  for  t  the  quartic 


3  3-2 

2/  -f"            p 

1  2 

-t^  t             ^ 


=  0; 


i.e.,  p'^4_|_2y/3_^2^2_|.2^^_|_^  =  0. 

This  shows,  as  before  determined,  that  there  are  at  most  4  cardioids 
satisfying  the  specified  conditions.  We  shall  now  seek  the  regions 
of  the  plane  where  there  are  4  real,  2  real,  or  none. 

Now,  for  a  quartic  (a,  6,  c,  6?,  e){x,  \Y  with  realc  oefficients, 
A  =  0  indicates  the  coincidence  of  2  of  the  4  roots; 
A<0  gives  2  real  and  2  imaginary  roots; 
A >0  gives  0  or  4  real  roots; 

A > 0  and,  besides,  62_ac>Oand  12(62— ac) 2- (ae—46<Z+3c2)a2>0 
gives  4  real  roots.* 

The  locus  A  =  0  will,  then,  separate  the  different  regions  of  the 


*  Cf.  Halphen,  Trait e  des  Fonctions  Elliptiques  et  Leurs  Applications,  Paris, 
1886,  Premiere  Partie,  p.  123. 


18  ,      Cardioids  Fulfilling  Certain  Assigned  Conditions 

plane  which  we  seek.  In  order  to  apply  these  criteria,  transform 
the  unit  circle  into  the  real  axis  by  the  transformation 

x-\-r 

by  which  the  quartic  becomes 

The  discriminant  of  this  quartic  is 

A=-256pp\p-j-p'-2)[27pp\p-\-p'-2)+8] 

12ff2_/a2  =  ^(2-.s'0- [(2-3^1)2-4] 

o 

The  discriminant,  equated  to  0  and  geometrically  interpreted, 
breaks  up  into  the  lines  0/,  OJ  (which,  being  imaginary,  may  be 
disregarded),  the  line  p-\-p'—2  =  0  and  the  circular  cubic 
27pp'{p-\-p'  —  2)-\-8  =  0.  Plotting  these,  we  can  easily  mark  the 
sought  regions  of  the  plane. 

For  points  on  the  cubic  and  its  asymptote,  A  =  0;  i.  e.,  there  are 
2  of  the  4  cardioids  coincident.  Note  that  p  given  in  such  a 
position  would  be  a  cusp  (on  the  cubic)  or  a  point  of  tangency  (on 
the  line).  For  points  not  on  these  curves,  A  is  either  >0  or  <0; 
hence,  these  curves  mark  off  the  regions  of  the  plane  whei*e  there 
are  4,  2,  or  0  real  curves. 

Let  us  test  points  in  the  various  sections.     For  point  _,  A>0 

and  12H^  —  Ia^<0.     Therefore,  for  points  in  the  region  where 

is  (i.  e.,  in  the  loop),  there  are  no  real  cardioids.  This  was  to  be 
expected,  for  0  is  the  given  centre  and  the  cubic  is  the  cusp-locus 
for  cardioids  touching  z-\-z'  =  2.  Therefore,  p  taken  between  the 
centre  and  the  cusp  would  certainly  lead  to  no  real  cardioid. 

For  point  2,  A  <  0,  which  indicates  that  p  in  the  region  exterior 
to  the  circular  cubic  and  its  asymptote  gives  2  real  and  2  imaginary 
cardioids. 

For  point  1  A>0,  b^-ac>Osind  12H^-Ia^>0;  therefore,  in  the 
region  between  the  cubic  and  its  asymptote,  p  gives  4  real  cardioids. 


Cardioids  Fulfilling  Certain  Assigned  Conditions        19 

Thus,  p  will  determine  the  number  of  real  cardioids  according 
to  its  position  in  the  various  regions  marked  in  Fig.  III. 

It  is  worthy  of  remark  that,  taking  0  as  the  given  point,  z-{-z'  =  2 
as  the  given  line  and  c  as  the  centre,  the  discriminant  of  the 
resulting  quartic  in  t  is  the  parabolic  cubic 

which  forms  the  centre-locus  for  cardioids  with  cusp  0  and  line 
2+2^  =  2.  Centres  in  the  loop  give  imaginary  solutions;  between 
the  infinite  branches  there  are  4  real  cardioids;  centres , at  other 
points  in  the  plane  yield  2  real  and  2  imaginary  curves. 


Fig.  hi. 

PROBLEM  (c) :  Given  the  centre  and  2  points. 

Take  the  centre  at  the  origin;  the  points  as  1  and  p. 

The  cusp-locus  for  cardioids  with  centre  0  and  passing  through 
1  is 

Sn'^+6kk' -4^kk'{k-\-k')  -1  =  0. 
The  cusp-locus  for  cardioids  with  centre  0  and  passing  through  p 
is 

3k^k''+6kk'pp'  -  ^kk'{kp'+k'p)  -  pV^  =  0. 
These  two  curves  have  16  common  intersections;  i.  e.,  there  should 
be  16  cardioids  satisfying  the  given  conditions.  But,  since  the 
two  curves  have  double  points  at  I  and  J  with  one  branch  of  each 
having  a  flex  thereat,  and  the  flex  tangents  also  being  common, 
there  are  12  intersections  at  /  and  /.  Therefore,  there  can  be,  at 
most,  4  real  cardioids.     Let  us  now  determine  the  regions  of  the 


20         Cardioids  Fulfilling  Certain  Assigned  Conditions 


plane  where  there  are  4,  2,  or  0  curves  satisfying  the  required 
conditions. 

From  the  conditions  that  the  sought  curves  have  centre  0  and 
pass  through  1  and  p,  we  obtain  the  equations : 

2at-a'f  =  p]  2aT-aV=l^ 


From  these  a,   a',   r  must  be  eHminated.     The  eHmination  of 
these  quantities  leads  to  the  quartic  in  t^ : 

4p'^{p'  -i)t'^+HSp'^-5pp''+Qpp"'  -9p'^)i^ +s(npY^  -  Wpy 

-  l6pp'^+S0pp'-9)t^-\-^(Sp^-5p'p^-i-6pY-9p^)t^ 
-\-4p^{p -1)  =  0. 


Fig.  IV. 
For  this  the  invariants  are 

41  =S^{S2-l){S2^-\-2lS2''-S2SiS2  +  51S2-9) 

8J  =  S^(-S2^-\-ms2^-H0s2^Si-Ss2^-\-4>8s2hi+108s2^+585s2^ 

-Q24>SiS2''+nSSih2^+U4>SiS2-270S2  +  27) 

'  -  26.925l3  +  2%l2(  -  252^+  \0S2' 

+  12^2-1) 
+  225i(-52^+ 18^2^-5452^- 128502 
-3352+6) 

-h(352«-3052^+1752'  +  30052'  +  21352' 
+  1852-9) 
=  3352^(45i  +  52--652-3)3  (-45152  +  352^  +  652-  1). 

451+522  —  652  —  3  =  0  is  the  cardioid  with  cusp  at  1,  centre  at  0; 
—  45152+352^+652- 1  =  0  is  the  bicircular  quartic  with  cusp  at  1 
which  is  the  locus  of  cusps  of  cardioids  with  centre  at  0  and 
passing  through  1.     It  would  seem  that  there  are  never  more 


A  =  3^52^451  +  522-652-3) 


Cardioids  Fulfilling  Certain  Assigned  Conditions        21 

than  2  cardioids  real,  for  if  the  second  point  lies  either  out^de  the 
cardioid 

451  +  52^-652-3  =  0, 

or  inside  the  quartic 

-45152  +  3522  +  652-1  =  0, 

there  will  be  none  real;  if  it  lies  between  these  curves,  there 

will  be  2  real. 

PROBLEM  (d) :  Given  the  cusp  and  2  lines. 

Take  the  cusp  at  0;  the  lines  as  2+2'  =  2  and  2  =  2:'^i+ai: 

The  centre-locus  for  cardioids  with  cusp  at  0  and  touching  the 
line  2+z'  =  2  is  the  parabolic  cubic 

(c+c'-2)3+54cc'  =  0. 
The  centre-locus  for  cardioids  with  cusp  at  0  and  touching  the 
line  z  =  z'ti-\-ai\s  the  parabolic  cubic 

(c-c'^i+ai)3-27cc'ai^i  =  0. 
These  two  curves  have  9  common  intersections.  By  a  repetition 
of  the  argument  used  in  Problem  (a),  it  may  be  shown  that  3  of 
these  are  always  real.  And,  as  a  matter  of  fact,  these  loci  have 
but  3  real  intersections;  i.  e.,  there  are  3  cardioids  with  a  given 
cusp  and  touching  2  lines.  See  Fig.  IX,  p.  43,  where  the  3  cardioids 
are  shown. 
PROBLEM  (e) :  Given  the  cusp,  a  line  and  a  point. 

Take  the  cusp  at  0;  the  line  as  z+z'  =  2,  and  the  point  as  p. 

The  locus  of  centres  of  cardioids  with  cusp  at  the  origin  and 
touching  the  line  2+2' =  2  is  the  nodal  cubic 
(c+c'-2)3+54cc'  =  0. 
The  centre-locus  for  cardioids  with  0  for  cusp  and  on  the  point  j) 
is  the  parabola 

^cc'fj)'  =  {cp'-]-c'p  —  pp'y. 
These  confocal  curves  have  6  common  intersections,  2  of  which 
seem  to  be  always  imaginary;  i.  e.,  there  seem  to  be  at  most  4  real 
cardioids  having  a  given  cusp,  passing  through  a  given  point  and 
touching  a  given  line. 
PROBLEM  (f) :  Given  the  cusp  and  2  points. 

Take  the  cusp  at  0;  the  points  as  1  and  p. 

The  centre-locus  for  cardioids  with  cusp  at  0  and  passing  through 

1  is  the  parabola 

4cc'=(c+c'-l)2. 


22         Cardioids  FuipUng  Certain  Assigned  Conditions 

The  ceptre-locus  for  cardioids  with  cusp  at  0  and  passing  through 
p  is  the  parabola 

4!cc'pp'  =(cp'-\-c'p  —  pp'y. 
These  conies  have  4  intersections  in  common;  but  being  confocal 
parabolas,   only   2  of  these   intersections   are  real.*     There  are, 
then,  only  2  real  cardioids  with  a  given  cusp  and  2  given  points 
(Fig.  XI). 

PROBLEM  (g) :  Given  the  double  tangent  and  2  lines. 

Take  the  double  tangent  as  z-{-z'  =  2;  the  lines  as  z  =  tiz'  and 
z  =  t2z'-\-a2. 

The  centre-locus  for  cardioids  with  double  tangent  z-\-z'  =  2  and 
touching  the  Hue  2  =  2^/1  consists  of  three  lines  through  the  inter- 
section of  2H-2'  =  2  and  z  =  z'ti  such  that  if  ti  =  e^^^,  the  inclinations 

of  the  lines  to  the  axis  of  reals  are  ^,  —^7—,  — - — ,  respectively. 

Similarly,  the  centre-locus  for  cardioids  with  2+2' =  2  as  double 
tangent  and  touching  z  =  t2z'-{-a2  consists  of  three  lines  through 
the   intersection   of   z-\-z'  =  '2   and    z  =  2^^2+02   with   inclinations 

o.  ~~^'  ""^ — »  respectively,  where  t2  =  e^i<P.     These  loci  intersect 

in  9  real  points;  so  there  are,  in  general,  9  cardioids  with  a  given 
double  tangent  and  2  lines.  If,  however,  the  2  given  lines  are 
parallel,  there  are  only  6  proper  cardioids;  and  if  1  line  is  parallel 
to  the  double  tangent,  there  are  but  3. 

PROBLEM  (h) :  Given  the  double  tangent,  a  point  and  a  line. 

Let  the  double  tangent  be  2+2' =  2,  0  the  given  point,  and 
2=  2'<iH-ai  the  given  line. 

The  centre-locus  for  cardioids  on  0  and  touching  2+ 2' =  2  as  a 
double  tangent  is  the  quartic: 

(2-2-20^  +  4(2-2-20' (2+20  +  18(2-2-20222'-272V2  =  O. 

The  centre-locus  for  cardioids  touching  2  =  2'/i+ai  and  having 
2+2' =  2  as  a  double  tangent  consists  of  the  3  lines  as  described  on 
page  13.  Since  a  double  tangent  and  a  centre  determine  a 
cardioid,  we  should  expect  the  number  of  cardioids  fulfilling  the 
required  conditions  to  be  12,  as  given  by  the  intersections  of  the 
quartic  and  the  -3  lines.  However,  because  of  the  inclination  of 
the  parabolic  branches  of  the  quartic  and  the  60°-inclination  of 

*  Cf.  Charlotte  Scott,  Modern  Analytical  Geometry,  N.  Y.,  1894,  p.  77. 


Cardioids  Fulfilling  Certain  Assigned  Conditions        23 

the  lines  towards  each  other,  there  are  always  at  least  2,  and 
possibly  4,  of  the  intersections  imaginary.  Hence,  there  are  at 
most  10,  and  possibly  only  8,  real  cardioids  with  a  given  point, 
a  given  double  tangent  and  touching  a  given  line. 

PROBLEM  (i) :  Given  the  double  tangent  and  2  points. 

As  a  preliminary  statement,  let  us  remark  that  the  2  points 
must  be  both  on  the  same  side  of  the  double  tangent;  otherwise, 
there  would  be  no  real  solution. 

Let  2+2;'  =  2  be  the  given  double  tangent;  0  and  p  the  given 
points. 

The  cusp-locus  for  cardioids  with  z-\-z'  =  2  as  double  tangent 
and  going  through  the  origin  is  the  quartic 

The  cusp-locus  for  cardioids  with  double  tangent  2+z'  =  2  and 
passing  through  p  is  the  quartic 

^2-k-k'y{k-p){k'-f)  =  [i2-k-k'){p'-k'+p-k) 

+{p-k){p'-k')Y 

These  two  curves  have  doubly  parabolic  branches  which  are 
respectively  parallel.  Moreover,  they  each  consist  of  a  cuspidal 
and  a  non-cuspidal  part,  which  we  may  designate  for  the  two 
curves  (Q,  Q^)  as  K,  N,  K^,  N^,  respectively. 

Since  a  double  tangent  and  a  cusp  uniquely  determine  a  cardioid, 
the  number  of  cardioids  satisfying  the  required  conditions  is 
given  by  the  number  of  intersections  of  the  quartics  Q  and  Q^. 
There  are  16  such  intersections.  Of  these,  however,  4  lead  to 
degenerate  curves  the  cusps  of  which  are  on  the  line  at  infinity  at 
the  points  of  intersection  of  the  two  pairs  of  parallel  parabolic 
branches.  Let  us  now  determine  how  many  of  the  12  remaining 
intersections  can  be  real. 

N  and  iV\  K  and  K^  can  each  intersect  but  once;  A^  and  K^,  K 
and  N^  may  have  0,  1,  or  2  intersections.  Moreover,  p  may  be  so 
placed  as  to  combine  the  maximum  number  of  intersections  of  the 
cuspidal  and  non-cuspidal  parts  of  the  respective  curves  (see 
Fig.  V,  p.  43).  Thus,  there  are,  at  most,  6  real  proper  cardioids 
going  through  2  points  and  touching  a  given  double  tangent. 

Here  conclude  the  simpler  problems,  in  each  of  which  the  data 
include  one  of  the  fixed  elements  of  the  cardioid.  We  shall  now 
consider  a  few  more  difficult  problems  in  which  lines  and  points 
form  the  data. 


More  Difficult  Problems. 

I.  When  3  Conditions  Are  Given. 
PROBLEM  (7) :  Given  3  lines;  find  the  locus  of  centres. 

Let  the  3  given  lines  be  z  =  z'ti-\-ai  {i=  1,2.3). 
Since  each  of  the  lines  is  to  be  tangent  to  the  cardioid,  we  have 
the  identities : 

z-z'P-zo-Sat-\-Sa'f-}-z'ot^--z-z%-a, 
whence,  t^  =  ti (1) 

and  Zo-\-Sat-a'f-z'ot^  =  ai (2) 

From  (1),  we  get  /  =  ^V  U  =  Ti,  cor,,  coV,-. 

Substituting  the  value  t  =  Ti  in  (2),  we  get  the  three  equations: 

Zo-^3aTi  —  Sa'Ti^  —  z'oTi^  =  ai 


zo+3aT2  — 3aV 


Zo-{-SaT2  —  Sa'Tz^  —  z'oTs^  =  a3 
Eliminating  a,  a'  from  these  3  equations,  we  obtain  the  line 

=  0,  as  part  of  the 

locus.  Since  we  can  choose  each  of  the  3  values  oft{t  =  Ti,  cor,,  co're) 
in  3  ways,  there  are  27  different  combinations  or  27  different  sets 
of  3  equations  in  zo,  z'o,  a,  a'.  But  these  lead  to  only  9  distinct 
lines,  arising  thus : 


Zo—z'oTi^  —  ai 

Tl 

Tl' 

Zo—z'oT'^  —  ai 

T2 

T2' 

Zo  —  z'oTz'  —  Ciz 

T3 

T3^ 

[ti,         r2,         T; 

{coTiy    cor 2,    CO 


The  combinations  ^cori,  cor 2,  cors  >  give  the  same  line,  which 

[coVi,  C0V2,  C0V3J 

we  shall  designate  as  Zi. 


coVi,  r2,      cor 3  j-  give  Zj 
n,      cor2,   C0V3J 


T2y         OiTs 


0)Tu 

cor  2, 

C0V3 

give  Z4; 

coVi, 

C0V2, 

r3     J 

TU 

cor2. 

T3       1 

cori. 

WV2, 

cor3 

■  give  Ze; 

coVi, 

T2y 

C0V3 

cori, 

T2y 

T3 

coVi, 

cor2, 

cor3 

>  giveZg; 

Tl, 

C0V2, 

C0V3 

24 

Cardioids  Fulfilling  Certain  Assigned  Conditions        25 

The  directions  of,  these  lines  are  determined  by  the  ratio  of  the 
coefficient  of  Zo  to  the  coefficient  of  z'o  in  the  respective  equations. 
The  coefficient  of  zo  for  Zi  is 


1         ri 

1  T2 

1  rs 


=  (ri  — r2)(T2  — r3)(T3  — n). 


The  coefficient  of  z'o  for  the  same  Hne  is 


73^  T3 


=  —TiT2T3(ti—T2){t2—T3){t3  —  Ti). 


The  ratio  of  the  coefficients  of  Zo  and  z'o  is  1:  — 0-3.  Likewise,  it 
will  be  found  for  L2  and  Z3  that  the  ratio  of  the  coefficients 
of  Zo  :  and  z'o  is  1:  —  0-3;  i.  e.,  Zi,  Z2,  -L3  are  parallel. 

Similarly,  Li,  Ze,  Ls  are  parallel  with  directions  determined 
by  the  ratio  1 :  — cocs;  and  Z5,  Lj,  Z9  are  also  parallel  with  directions 
determined  by  1:  — C0V3.  Now,  if  6,  cp,  \f/  be  the  angles  these 
t  hree  sets  of  lines  make  with  the  real  axis,  then 

Similarly,  0—4/==^=- 

o 

and  <p—^==^^\ 

i.  e.,  the  sets  of  parallel  lines  are  inclined  at  angles  of  60°  towards 
one  another. 

Thus,  the  locus  of  centres  of  cardioids  touching  3  given  lines 
consists  of  3  sets  of  parallel  lines,  inclined  to  each  other  at  angles 
of  60°.     (See  Fig.  VI.) 

Let  us  endeavor  to  interpret  this  locus  geometrically.  Desig- 
nating the  3  given  lines  as  h,  h,  h,  respectively,  of  which  Zi  and  Zj 
intersect  at  an  angle  a,  let  us  consider  any  one  cardioid  touching 
these  given  lines.  This  cardioid  possesses  a  cusp  which  may 
have  any  one  of  three  positions  relative  to  Zi  and  Z2  (its  relation  to 
Zs  being,  for  the  present,  left  out  of  account)  according  as  the 

cuspidal  rays  to  the  points  of  tangency  make  an  angle  of     a, 

•1  o 


2G         Cardioids  Fiil filling  Certain  Assigtied  Conditions 

2  2 

Qa  +  120°,  or  ^aH- 240°.     With  each  of  these  three  cusps  is  asso- 

o  o 

ciated  a  definite  centre  which,  as  the  variable  cardioid  assumes 
positions  fulfilHng  the  required  conditions,  traces  out  one  of  the 
nine  lines  found,  as  the  centre  locus.  Moreover,  each  of  these 
three  lines  forms  an  angle  of  120°  (or  60°)  with  the  others.  Further, 
combining  each  of  these  three  positions  relative  to  Zi,  1 2  with  the 
three  possible  relations  with  Zs,  we  obtain  the  complete  locus  of  9 
lines  as  above. 

PROBLEM  (8):  Given  2  lines  and  a  point;  find  the  locus  of 
centres. 

Let     \       z'      \he   the   given   lines,   and   p   the   given   point. 

1^=7?  J 

The  identification  of  these  lines  with  a  tangent  to  the  cardioid 
leads  to  the  equations : 

111 

/'3=,/^2.   t'  =  t^\  Ojti\  Oi%^ (1) 

Zo-z'ot''-\-Sai'-Sar'~  =  0 (2) 

T'  =  ~-,;r  =  ti    ^a^/l    \icHi    3 (3^ 

2o-2V+3ar-3aV  =  0 (4) 

We  have,  also,  the  equations : 

p-zo-^at+a't^  =  0 (5) 

(z'o-py^+^a't-a^O (6) 

The  elimination  of  Z,  Z',  r,  a,  a'  from  these  six  equations  will  give 
the  required  locus. 

From  (2),  (4),  we  obtain 

—  ZoSi  —  z'oS'^ 

3a  =  - 


From  (5),  (6): 


where  r^  =  ;;+^ 

—  Z0  —  Z0S1S2  \S2  =  tT 


a'  —2a  p  —  zo  0 

0  a'  —2a  v  —  zo 

z'o-p'  2a'  -a  0 

Q  z'o-p'  2a'  -a 


0. 


Cardioids  Fulfilling  Certain  Assigned  Conditions        27 

Substitution  of  the  values  of  a,  a'  in  this  determinant  furnishes  a 
quartic  as  part  of  the  sought  locus.  Since  there  are  3  values  each 
for  t'  and  r,  there  seem  to  be  9  quartics  in  Zo,  z'o,  ^i,  ^2.  But  these 
9  reduce  to  8;  for  the  combinations  of  t\  r,  divide  off  in  sets  of  3, 
producing  only  3  distinct  developments  of  the  determinant. 
These  combinations  are,  in  fact,  the  following: 


t\      r      1  Oil',   coV  ]  oiH\ 

a)t',    cor    \  ,  coH',  r       >  ,  t'. 


H\  coV  r,      o)T  cor, 


cor 
coV 


r 


Now  the  only  terms  in  the  development  involving  a  or  a'  are  those 
in  a^,  a'^  aa\  a^a'^.  The  effect  of  replacing  t\  r  by  cat',  cor,  respec- 
tively, in  the  values  of  a  and  a'  is  to  multiply  the  equation  by 
co^(  =  l).  Similarly,  the  substitution  of  co^^',  coV  for  t'  and  r, 
respectively,  multiplies  the  equation  by  1.  Thus,  the  combina- 
tions in  the  first  set  lead  to  the  same  line.  Likewise,  those  in  the 
other  2  sets  lead  to  but  2  other  distinct  lines. 

These  3  sets  of  combinations  lead  to  the  following  values  of  ^i,  ^2: 

2 


Sl  = 

=  2  cos 

i« 

Si- 

=  1 

Sl= 

=  2  cos 

(l-H 

52  = 

=  1 

5i  = 

=  2  cos 

(M') 

52  = 

=  1 

where  ti  =  e  .  Hence,  52  may  always  be  taken  equal  to  1,  and 
the  development  is 

iZo  +  z'oSiy{ZoSi  +  z'oy-4[{Zo-p){Zo  +  z'Siy-{-{z'o-p'}(ZoSi  +  z'oy] 

-\-18{zo-p)iz'o-p')izo-\-z'oSi)(zoSi-\-z'o)-27{zo-pylz'o-p'y  =  0, 
where  Si  has  one  of  the  values  above.  This  equation  can  be  put 
in  the  form 

(si''-4^)W-SiZoz'o-\-z'o^y+f(zo,  z'o,  p,  py  =  0, 
which  indicates  2  parabolic  branches.     These  branches  are  inclined 

at  angles  ±-,  =^  I  0+^  )'  ^  (  q"^  q  I  ^^^  ^^^  three  quartics,  respec- 
tively. Moreover,  the  locus  has  a  cusp  at  the  point  where  the 
variable  cardioid  comes  into  the  position  where  the  fixed  point 


28         Cardioids  Fulfilling  Certain  Assigned  Conditions 

is  its  cusp.  Thus,  the  centre  locus  consists  of  3  quartics  with 
properties  as  described  above.  The  curves  are  of  the  form  shown 
in  Fig.  VI. 

Geometrically  interpreted,  it  appears  that  the  3  quartics  are 
traced  out  by  the  centres  of  those  cardioids  (touching  the  given 
lines  and  on  the  given  point)  of  which  the  cuspidal  rays  to  the 

points  of  tangency  are  inclined  towards  each  other  at  angles  -9y 

o 

2  2 

-^+120°,  -^+240°,  respectively,  where  0  is  the  angle  between  the 

o  o 

given  lines. 

The  specialization  of  this  problem,  arising  when  p  is  the  point 
of  tangency  on  one  of  the  given  lines,  proves  instructive.  Let  uS 
first  view  the  problem  from  a  geometric  standpoint  and  see  what 
a  priori  information  we  can  derive  therefrom.  Consider  a  case 
where  3  lines,  lu  h,  h,  are  given,  of  which  h  and  I2  are  fixed  in 
position,  while  /s  rotates  about  a  fixed  point  p  on  h.  For  any 
position  of  the  variable  line,  the  centre-locus  consists  of  3  sets  of 

3  lines  as  found  in  Problem  (7).  The  limiting  position  of  h  is 
that  in  which  it  comes  into  coincidence  with  li.  Then  p  becomes 
the  point  of  tangency  on  /i,  and  the  conditions  become  equivalent 
to  the  data  of  this  specialized  problem.  Thus,  we  may  expect 
sets  of  lines  in  the  centre-locus.  Will  the  3  sets  appear.^  To 
answer  this,  recall  the  fact  that  the  3  sets  of  lines  in  the  centre- 
locus  of  Problem  (7)  are  associated  with  cardioids  such  that  the 
cuspidal  rays  to  the  points  of  tangency  make  angles  equal  to 

2     2  2 

-dy  -^+120°,  -^-f  240°,  respectively,  where  0  is  the  angle  between 

the  tangents.  In  the  case  under  consideration  6,  the  angle  between 
li  and  /s,  is  0;  and  the  cuspidal  rays  to  the  point  of  tangency  make 
an  angle  0°  only.  Thus,  the  3  sets  of  3  lines  seem  to  reduce  to 
one  set  of  3.  What  happens  to  the  other  2  sets  of  lines  .^  They 
are  evidently  associated  with  cardioids  such  that  the  cuspidal 
rays  to  the  points  of  tangency  make  angles  of  120°  and  240°, 
respectively;  i.  e.,  with  cardioids  of  which  the  tangents  make 
angles  of  180°  or  360°,  that  is,  with  cardioids  with  parallel  tangents. 
Thus,  the  other  2  sets  of  lines  belong,  not  to  the  case  where  Iz 
and  li  are  coincident,  but  to  the  case  where  they  are  parallel.* 

*  Cf.  the  theorem  indicated  on  p.  8. 


Cardioids  Fulfilling  Certain  Assigned  Conditions        29 
Let  us  now  consider  the  problem  analytically. 

liCt  us  take  ]  „_^_    r  as  the  given  lines,  and  p  as  the  point  on 

[        ti"    I 
z  =  t^z'.     Let  us  find  the  locus  of  centres. 

The  identification  of  the  given  lines  with  a  tangent  leads  to 
equations  (1),  (2),  (3),  (4)  of  Problem  (8).  Since  p  is  the  point 
of  tangency  on  z  =  tiz'  and  is  also  on  the  variable  cardioid,  we  have 

p  =  Zo^%at'-a't'^ (5) 

The  elimination  of  a,  a'  from  (2),  (4)  and  (5)  gives  the  line 

Zo-z'or^         St  3r2 

Zo-z'qI'^     '    St'  St""         =0 

Zo-p  n'  t"" 

as  the  required  locus. 

By  development  of  the  determinant,  this  line  is  found  to  be 

Zo{^t'\  - i'T"- - 1'^)  -  z'o  t'^-ri^i'^  - t'T"- - 1'^)  -  Spt'rif  -  r)  =  0. 

Although  there  are  9  possible  combinations  of  t'  and  r,  there  are 

not  the  same  number  of  lines  in  the  locus.     In  fact,  since  t'  =  -^ 

T 
2 

",  or  -  ,  and  the  combinations  of  t'  and  r  are  all  of  the  third 

r  T 

2  2  t_ 

degree,  there  are  only  3  distinct  lines  with  clinants  ^i^,  co^l^  a)^<l^ 
respectively;  i.  e.,  the  3  lines  are  inclined  towards  each  other  at 
angles  of  60°  (or  120°). 

PROBLEM  (9):  Given  2  points  and  1  line;  find  the  locus  of 
centres. 

Take  the  line  as  z-\-z'  =  %,  the  points  as  0  and  p.  Let  the  equa- 
tion of  the  cardioid  be 

z  =  zo-\-^at  —  a't^. 
Since  the  cardioid  touches  2+2;' =  2,  we  have  the  identity: 
z-zo-  iz' -z' o)t^-Sat-Sa't'^  ~  z-\-z' -'l\ 

whence,  /^=  —  1;  ^=  —  1,  -co,  — co^ (1) 

zo-z'ot^-^Sat-Sa'f  =  % (2) 

Since  p  and  0  are  points  on  the  cardioid,  we  have 

p  =  Zo+^at'-a't'^ .  (3) 

f      ,    ,  ^a'     a  , 

V  =^o+Y  ~j2 W 


-a' 

2a 

Zo  —  p 

0 

0 

-a' 

2a 

Zo 

p'-z'o 

^^a' 

a 

0 

0 

v'-z'o 

-2a' 

a 

30         Cardioids  Fulfilling  Certain  Assigned  Conditions 

0  =  2o+2ar-aV2 (5) 

o=,'„+?£'_«^.. ..(e) 

The  elimination  of  t,  t\  r,  a,  a'  from  these  6  equations  will  give 
the  required  centre-locus. 

Eliminating  t'  from  (3),  (4),  we  obtain  the  equation: 


=  0. 


Developed,  this  is 

'Sahi'^-^a\z'o-p')-^a'\zo-v)-Qaa'{zo-v){z'o-p') 

-i-{zo-py(z'o-pr  =  0 (7) 

The  elimination  of  r  from  (5),  (6)  leads  to 

3a2a'2-4a32o-4a'Vo-6aa'zo2'o+2oVo2  =  0 (8) 

From  (1),  employing  the  value  ^=  —  1, 

2o+2'o-3a-3a'  =  2 (9) 

From  (7),  (8),  (9),  a,  a'  have  still  to  be  ehminated.  Let  us 
transform  to  Cartesian  co-ordinates  by  replacing  a  by  a  +  i6, 
z  by  x-\-iyy  p  by  p-\-iq,  with  a  corresponding  substitution  for  their 
conjugates.     Equation  (8)  becomes 

Equation   (7)   becomes  a  similar  equation  in  which  x  —  py  y  —  q 

X—  1 
replace   x    and    i/,    respectively.     Equation    (9)    gives    a  =  — — . 

o 

Substituting  the  value  of  a  in  (7)  and  (8),  we  obtain  the  quartics 

in  6: 

2764-7263?/-M862(4;r2-9i/2-2x-n)+7262/(x- 1)2+/(.T,  ?/)4  =  0 

2764-726"X2/-^)  +  186y(x,  y,  p,  qy-\-72b{y-q){x-  ly 
-\-^{x,y,p,q)'^  =  0 

The  elimination  of  b  leads  to  an  8-row  determinant,  which,  devel- 
oped, is  an  equation  of  the  12th  degree  in  x  and  y.  Thus,  the 
centre-locus  is  a  12-ic.  We  shall  not  attempt  to  discuss  the 
singularities  of  this  locus. 

Let  us  now  find  the  centre-locus  for  the  case  where  one  of  the 
points  is  on  the  line;  that  is,  let  the  data  be  the  line  2+2'  =  2  and 
the  points  0  and  p,  of  which  p  is  on  the  given  line. 


Cardioids  Fulftlling  Certain  As^gned  Conditions        31 

From  p.  29,  we  have  /  =  —  1,  -co,  —  co^ (1) 

Zo-z'ot^-\-Sat-3a't'^  =  2 (2) 

SaW^-^a^zo-W^z'  o-Qaa'zoz'  o  +zo^z'o^  =  0 (3) 

Since  p  is  the  point  of  tangency  on  the  cardioid, 

j)  =  Zo-\-^at-a'f .(4) 

From  (2)  and  (4), 

3a/  =  3j9-22o+2V_2 

^'  =  3p'-2.^+f3-2 

Since  a,  a',  occur  in  (3)  only  in  the  combinations  a^a'^,  a\  a'^  aa\ 
any  of  the  values  /=  —  1,  —  co,  or  —  oj^  will  lead  to  the  same  result. 
Employing  any  one  of  these  values,  we  obtain  a  quartic  in  Zo,  z'o 
as  the  sought  centre-locus. 

We  might  have  expected  such  a  result,  for  the  data  from  the 
limiting  case  for  2  lines  and  a  point,*  where  the  lines  coincide  and 
their  point  of  intersection  becomes  the  point  of  tangency.  More- 
over, we  can  account  for  the  reduction  in  the  number  of  quartics 
in  the  locus  by  recognizing  that  the  2  quartics  which  do  not 
appear  here  belong,  not  to  the  case  where  the  2  lines  coincide, 
but  to  the  case  where  they  are  parallel. 

PROBLEM  (10):  Given  3  points;  find  the  locus  of  cusps. 

Let  the  given  points  be  0,  p,  q,  and  the  map-equation  of  the 
cardioid  be 

z  =  h-a{l-t)\ 

If  /i  be  the  value  of  ^  at  0, 

k-a{\-tiY  =  0\ 

whence,  {k¥-a'k-aky  =  4>aa'kk' (1) 

Similarly,  since  the  cardioid  is  on  ^, 

[(k-p){k'-p')-a\k-p)-a{k'-p')Y  =  4>aa\k-p)(k'-p').{2) 
Since  g-  is  a  third  point  on  the  cardioid,  we  have 

[{k-q){k'-q')-a'ik-q)-a{k'-q')]^  =  4>aa\k-q)(k'-q')..{S) 
We  have  to  eliminate  a,  a\  from  (1),  (2),  (3). 

*  Cf.  Problem  (8),  p.  26- 


32        Cardioids  Fulfilling  Certain  Assigned  Conditions 

Transforming  to  Cartesian  co-ordinates  and  letting 

k  =  X-\-iY 
a  =  A-^iB 

q=U-^iV 
the  equations  become 

(Z2+F2-2^Z-25F)2  =  4  (^2_^52)(Z2+r») (1) 

[{X-PY+{Y-QY-<iA{X-P)-W  {Y-Q)f 

=  4(.42+52)  [(X-P)2+(F-Q)2].  .  .(2) 
[{X-UY-V{Y-VY-%A{X-U)-W  {Y-V)f 

=  4(^2+^2)  i(^x-UY+{Y-'Vy]. .  .(3) 

Interpreting  A  and  B  as  Cartesian  co-ordinates,  these  equations 
represent  parabolas  with  0  as  focus.  We  seek  the  condition  that 
the  three  curves  have  a  common  point. 

Since  (1)  and  (2)  [also  (1)  and  (3)]  are  confocal  parabolas, 
they  have  3  known  common  lines,  of  which  two  are  imaginary  (the 
lines  from  I  and  J);  and  the  third  (the  line  at  infinity),  real. 
Their  fourth  common  line  is,  therefore,  rationally  obtainable  and, 
therefore,  their  self -con  jugate  triangle.  This  triangle  will  have 
1  real  vertex  and  2  imaginary  ones;  and  the  one  real  pair  of  common 
chords  of  the  two  parabolas  will  pass  through  the  real  vertex. 
If  we  find  the  equations  of  the  line-pairs  common  to  parabolas  ( 1) 
and  (2),  (1)  and  (3),  respectively,  and  require  that,  with  properly 
chosen  sign,  they  be  simultaneously  true,  their  solution  will  yield 
the  co-ordinates  of  the  point  which  is  to  be  common  to  the  3 
parabolas.  The  substitution  of  these  values  in  the  equation  of 
one  of  the  parabolas  will  give  the  cusp-locus  sought. 

Parabola  (1)  in  line  co-ordinates  (using  ^,  ?;,  f  corresponding 
to  a,  a'y  w)  \^ 

V^k'+nk+^vkk'  =  0. 
Parabola  (2)  is 

Vnk'-p')+mk-p)+^rj{k-p){k'-p')  =  0. 

Their  fourth  common  line  is 

^ ^ I =  0 

kp\k-p)     k'p{k'-p')     kp'-k'p 

The  one  real  diagonal  point  of  the  4-line  is  [kp'{k  —  p)y  —k'p{k'—p'), 
kp'-k'p].  We  can  now  find  the  equation  of  the  sought  line-pair 
by  writing  the  pencil  C7+XF=0  built  up  on  the  2  parabolas,  and 


Cardioids  Fulfilling  Certain  Assigned  Conditions        33 


determine  X  so  that  the  resulting  curve  shall  contain  this  point. 
We  easily  find  as  the  equation  of  the  line-pair 

{k-v){k'-p'){W-ay-a'kY  =  kk\{k-p){k'-'p')-a{¥-'p') 

-a\k-j>)? 

Using  Cartesians,  the  line-pair  is 

[(X-P)2+(F-Q)2  [(X2-f-F2-2.4X-25F)2 


=  (Z2+F2)[X-P  +F-e  -^AX-P-WY-Q]\ 

X2-fF2  =  p2  '  ] 

Now,  if  we  let ^     2  I'  ,where  p,  g  are  taken    to   be 

X-P-\-Y-Q=a'] 

the  positive  square  roots,  the  2  lines  are 

(r(p2-2ylX-25F)=±p[(72-2^(X-P)-2B(F-Q)]....(4) 
Similarly,  the  line-pair  common  to  the  parabolas  (1)  and  (3)  is 

T(p2-2^X-2PF)  =  ^p[r2-2.4(Z-C/)-2P(F-F)]....(5) 
where  r  =  +^I{X-UY  +  {Y -Vy 

If  the  three  curves  are  to  have  a  common  point,  equations  (4)  and 
(5),  with  properly  chosen  sign,  must  be  simultaneously  true. 
In  terms  of  A  and  B,  these  equations  are 

2^[Pp-Z(a-+p)]-f2P[Qp-F(a+p)]  +  crp((7+p)  =  0.  .  .  .  (4) 
~    ^A[Up-X{T+p)]  +  ^B[Vp-Y{r+p)]^Tp{r+p)  =  Q.  .  ..(5) 

The  solution  of  these  two  equations  and  the  substitution  of  the 
resulting  values  of  A  and  B  in  the  equation  of  one  of  the  parabolas 
will  yield  the  cusp-locus  sought. 
From  (4),  (5) 


o"(o-+p) 
T{r+p) 


[Qp-Y{cT+p)] 
[Vp-Y{r+p)] 


[Pp-X{a-\-p)]    [Qp-F(cr+p)[- 
[Up-X{r-\-p)]    [Vp-Y{r-^p)] 


B 


(t{(T-\-p) 
r{r-\-p) 


[Pp-X{a^p)] 
[Up-X{r+p)] 


[Pp-X(cr-l-p)]    [Qp-Y{cr-\-p)] 
[Up-X{r+p)]    [Vp-Y{T^-p)] 


M         Cardioids  Fulfilling  Certain  Assigned  Conditions 


Designating  the  determinants  in  the  numerators  as  Di  and  D2, 
respectively,  and  the  denominator  determinant  as  Do, 

r  (/)i^+/)2^) 


AX-\-Br  = 


4        Do^ 

2  Do 


Di^+D2'  =  pV{p+,anU'  +  V^)  +  {a-Tnp-{-a)Hp  +  ry 

+  r'{p  +  rnp-^+Q^)-2aT{p+<T){p  +  T)(PU-hQV)] 
-2p((r+p)(pH-r)(cr-T)[(T(p+o-)(rZ+FF)-r(p+r)(PX+QF)] 


Z)lZ  +  2>2F  =  p     -(7(p  +  (7) 


X    Y 
U     V 


+  r(p-{-T) 


X 

p 


Do  may  be  expressed  thus 

P    Q 
u   V 


p\p 


X     Y  \ 


X    Y 

P    Q 


The  substitution  of  these  values  in 

[X'-\-Y'-2(AX-\-BY)Y  =  4(A'-\-B^){X'+Y^) 

gives  a  result  which,  after  division  by  p-,  is  of  the  form 

E-\-Fpa  +  GpT+HaT  =  0, 

where  E  is  of  the  sixth  degree  in  X,  F;  and  F,  G,  and  H  are  of  the 
fourth  degree  in  the  same  variables.     Rationalizing: 

{E+H(TTy=iFpa-\-GpTy 

E^-\-H^(t^t''-F'p^(t''-GVt^=(^FGp'--  2EH)(tt 

Squaring  again, 

E^+FY(T'+GYr'+H'(T^T^-2lE^F^(TV-\-E''GVT^-{-H^GYa^T' 
+E2HVt2+F202pV2t2+F277VVV]  +SEFGHp^a^T^  =  0 

Thus,  it  appears  that  the  required  locus  is  of  the  24th  degree. 
We  shall  not  attempt  to  discuss  the  singularities  of  the  curve 
beyond  remarking  that,  since  there  are  4  cardioids  with  a  given 
cusp  and  on  2  given  points,  it  would  seem  that  this  locus  has 
quadruple  points  at  each  of  the  3  given  points. 

Further,  we  may  reduce  the  unrationalized  equation  to  a  con- 
venient form  -thus : 
X^+Y'-2(AX-\-BY)  = 


p  {p' 

P  Q 
V  V 

-ip'-"') 

X  Y 
U  V 

+  {P'-T^) 

X  Y 

P  Q 

p 

P  Q 

U    V 

-(p+o) 

X  Y 

U   V 

+  {p+r) 

X   Y 

P  Q 

Cardioids  Fulfilling  Certain  Assigned  Conditions        35 


This  numerator  can  be  reduced  to 


-P 


—  P 


0 


P' 
p'+q' 

M2  +  r2 
X 

V 
u 
0 


,  which,  put  in  the  form 


and  equated  to  0,  is  the  product  of 


the  equations  of  the  circle  through  the  three  given  points,  0,  p^  q, 
and  the  hues  0/,  OJ.  Let  the  numerator  be  called  —pC.  Notin  a 
that  p,  0-,  r  are  factors  of  A'^-j-B^,  this  expression  may  be  designated 
as  p(7tK.     Thus,  the  required  locus  may  be  put  in  the  form 

-C  =  ^P(ttKDo, 

which  shows  that  the  locus  intersects  each  of  the  lines  joining  the 
given  points  with  I  and  J  only  at  these  circular  points. 

II.  When  4  Conditions  Are  Given. 

We  may  remark  that  in  the  problems  which  follow,  the  inter- 
sections of  loci  do  not  furnish  the  exact  number  of  solutions  as 
is  the  case  in  the  simpler  problems.  In  these  latter,  the  elements 
common  to  curves  on  both  loci  are  such  as  to  uniquely  determine  a 
cardioid;  while  in  the  cases  about  to  be  considered,  the  common 
elements  do  not  so  determine  a  single  cardioid. 

PROBLEM  (j) :  Given  4  tangents. 

Let  the  4  given  lines  be  z  =  z'U-\-at  {i=l,  2,  3,  4).  The 
centre-locus  for  cardioids  touching  z  =  z'ti-{-ai  (i=l,  2,  3)  con- 
sists of  the  9  lines  as  described  on  page  25.  The  centre-locus 
for  cardioids  touching  z  =  z'ti-{-ai  (i=l,  2,  4)  consists,  similarly, 
of  9  lines  which  divide  off  into  3  sets  of  parallel  lines  with  directions 
60°  apart. 

It  might  at  first  sight  seem  that  there  are  81  cardioids  touching 
the  four  given  lines.  But,  recalling  the  fact  that  the  locus  is  in 
each  case  generated  by  the  centres  of  3  variable  cardioids  so 
related  to  the  lines  h,  h  that  the  angles  between  the  cuspidal  rays 

2     2  2 

to  the  points  of  tangency  are    6,  _^+120°,  -^+240°,  respectively, 

it  is  readily  seen  that  only  the  intersections  of  corresponding  lines 
of  the  two  loci  give  a  common  centre  for  the  4  lines.     The  lines 


36         Cardioids  Fulfilling  Certain  Assigned  Conditions 

correspond  in  sets  of  three.    Thus,  there  are  27*  cardioids  touching 
4  given  lines. 

As  a  confirmation  of  this  result,  we  may  note  that,  although 
the  centre-loci  for  cardioids  touching  /i,  1 2,  h  and  h,  1 2,  h  intersect 
in  81  points,  these  intersections  will  not  give  81  cardioids  unless 
there  is  but  a  single  cardioid  with  a  given  centre  and  touching  2 
given  lines.  There  are  actually  3  such  cardioids,  which  suggests 
the  reduction  of  the  number  of  cardioids  touching  4  given  lines 
to  27. 

PROBLEM  (k) :  Given  3  lines  and  a  point. 

Let  the  given  lines  be  z  =  z'ti-{-ai  (z  =  l,   2,  3),  where  1^  =  - 

ti 

and  ai  =  a2  =  0;  take  p  as  the  given  point. 

The  centre-locus  for  cardioids  touching  z  =  z'ti-]-ai  (z=l,  2,  3) 
consists  of  three  sets  of  parallel  lines  similar  to  those  described 
in  Problem  (7).  The  centre-locus  for  cardioids  touching 
z=z'ti-{-ai  (i=  1,  2)  and  passing  through  p  consists  of  the  3  quar- 
tics  discussed  on  page  27. 

Although  these  loci  intersect  in  108  points,  there  are,  by  no 
means,  that  many  cardioids.  For,  with  each  centre  on  these  loci 
is  associated  a  definite  cusp;  and  it  is  only  the  intersections  of 
centre-loci  corresponding  to  the  same  cusp  that  give  a  common 
centre  for  a  cardioid  on  the  given  point  and  touching  the  given 
lines.  Now  these  loci  pair  off  in  such  a  way  that  to  each  quartic 
correspond  3  of  the  9  lines  ;t  so  that  there  are  36  centres  of  cardioids, 
and,  therefore,  36  cardioids  satisfying  the  given  conditions.  Fig.  VI 
illustrates  the  case  where  the  given  lines  form  an  equilateral 
triangle. 

When  the  point  is  on  one  of  the  lines,  there  are  but  9  cardioids. 
For,  designating  the  lines  as  lu  I2,  h,  with  ^  on  Zi,  the  centre-locus 
for  cardioids  on  Zi,  pi,  I2  consists  of  3  lines;t  likewise,  the  centre- 
locus  for  cardioids  onlu  pu  h  consists  of  3  lines.  These  loci  have 
9  common  intersections,  which  yield  centres  for  cardioids  on 
lu  hy  h  and  p. 


*  Professor  Morley  has  indicated  the  number  as  2^.  Cf.  Tram.  Amtr. 
Math.  Soc,  1900,  Vol.  1,  p.  114.  As  above  shown,  it  would  seem  that  there 
are  SK 

t  Cf.  p.  24. 

t  Cf.  Problem  (8),  p.  26. 


Cardioids  Fulfilling  Certain  Assigned  Conditions        37 


Fig.  VI.— The  loci  marked 


are  generated  by  the  centers  of 


cardioids  of  which  the  cuspidal  rays  to  the  points  of  tangency  make  angles 
of  40°,  160°,  and  280°,  respectively.  3  of  the  36  possible  cardioids  are  here 
shown.  For  this  particular  case,  there  are  actually  12  of  the  cardioids  real, 
as  shown  by  the  marked  centers.  % 

PROBLEM  (1) :  Given  2  lines  and  2  points. 

Let  the  given  lines  be  z  =  t^z'  and  z  =  —  ;  the  given  points,  p  and  q. 


The  centre-locus  for  cardioids  on  p  and  touching  z  =  t^z\  z  =  —^ 


38         Cardioids'  Fulfilling  Certain  A.^f^igned  Conditions 

consists  of  the  3  quartics  discussed  on  page  27.  The  centre- 
locus  for  cardioids  on  q  and  touching  the  given  lines  consists  of 
3  quartics  similar  to  the  above,  in  the  equations  of  which  q  and  q' 
replace  p  and  /)',  respectively. 

These  quartics,  which  we  may  designate  as  Qu  Q2,  Qs,  Q\,  Q't, 
Q's  for  the  two  loci  respectively,  pair  off  in  such  a  way  that  every 
intersection  of  Qi  and  Q'i  gives  a  cardioid  on  the  given  lines  and 
points.  There  are  48  intersections.  But  Qi  and  Q'»  are  both 
doubly  parabolic  with  the  same  points  at  infinity.  This  accounts 
for  4X3  intersections,  which  are  to  be  regarded  as  improper. 
This  leaves  36  proper  intersections;  i.  e.,  there  are  36  cardioids 
which  touch  2  given  lines  and  pass  through  2  given  points. 

If  one  of  the  points  is  on  one  of  the  lines,  this  number  reduces  to 
12.  For  the  centre-locus  for  cardioids  on /i,  h,  p\  (on  U)  consists 
of  3  lines;  the  centre-locus  for  cardioids  on  /i,  j>i,  7? 2  is  a  quartic. 
These  loci  have  12  common  intersections,  which  yield  12  centres 
for  cardioids  on  /i,  h,  pi,  />2- 

Further,  if  />2  is  on  1 2,  the  number  reduces  to  6.  For  the  centre- 
locus  for  cardioids  on  /i,  pi,  1 2,  consists  of  3  lines,  as  does  that  for 
cardioids  on  /i,  I2,  P2-  Although  these  loci  have  9  common  inter- 
sections, only  6  are  in  the  finite  plane;  for  since  the  clinants  of  the 
lines  in  both  loci  are  determined  by  the  angle  of  intersection  of 
li  and  /o,*  3  6f  the  intersections  are  at  infinity. 

PROBLEM  (m) :  Given  1  line  and  3  points. 

Let  the  given  line  be  2+z'  =  2;  the  given  points,  0,  p,  q.  The 
centre-locus  for  cardioids  on  0  and  p  and  touching  2-f  z'  =  2  is 
a  curve  of  the  12th  degree;  the  centre-locus  for  cardioids  on 
0,  qy  and  the  given  line  is,  likewise,  a  curve  of  the  12th  degree. 
These  curves  have  144  common  intersections;  so  there  cannot 
be  more  than  144  cardioids  which  touch  the  given  line  and 
pass  through  0,  p,  and  q.  There  will  likely  be  many  less  than 
144;  and  the  fact  that  there  are  4  cardioids  with  a  given 
centre,  point  and  line  suggests  that  this  number  may  reduce  to  36. 

If  one  of  the  points  is  on  the  given  line,  there  is  a  further  reduc- 
tion of  the  number  to  12.  For,  the  centres  of  cardioids  on  /i,  0, 
and  p  (where  p  is  on  h)  describe  a  quartic;  the  centres  of  cardioids 
on  /i,  p,  q,  describe  a  quartic.  These  loci  have  16  common  inter- 
sections, only  12  of  which  yield  centres  of  non-degenerate  curves. 

»  Cf.  p.  29. 


Cardioids  Fulfilling  Certain  Assigned  Conditions        39 

PROBLEM  (n) :  Given  4  points. 

Let  0,  p,  q,  r  be  the  given  points. 

The  cusp-locus  for  cardioids  on  the  points  0,  p,  q  is  a  curve  of  the 
24th  degree;  similarly,  the  cusp-locus  for  cardioids  on  0,  p,  r  is  a 
curve  of  the  same  degree.  Although  these  two  loci  intersect  in 
24^  points,  there  will,  by  no  means,  be  24^  cardioids  on  the  4  given 
points.  While  we  shall  hot  attempt  to  ascertain  the  exact  number, 
we  remark  that  the  fact  that  there  are  4  cardioids  with  a  given 
cusp  and  on  2  given  points  will  greatly  reduce  this  24^. 

We  shall  conclude  with  a  brief  discussion  of  a  problem  related 
to  those  we  have  been  treating,  the  solution  of  which  would  prob- 
ably prove  of  value  in  the  determination  of  the  number  of  cardioids 
on  4  points.  It  has  to  do  with  the  number  of  real  equilateral 
triangles  which  can  be  inscribed  in  a  given  cardioid  when  1  vertex 
is  fixed. 

Take  the  given  cardioid  with  its  cusp  at  0,  and  at  the  right  of 
the  figure.  Its  map-equation  is  z=  —{l  —  ty.  Let  this  be  the 
standard  size. 

First,  let  us  consider  a  triangle  of  any  shape,  and  let  us  determine 
the  relations  between  the  vectors  to  the  vertices.  Let  the  vectors 
to  the  vertices  of  a  triangle,  of  which  the  sides  and  angles  are 
Pi,  P2,  ps,  <p,  ^y  0,  respectively,  be  a,  6,  c,  the  terminal  end  of  vector 
a  being  at  the  apex  of  the  angle  <^,  of  vector  b  at  xj/,  and  of  vector 
c  at  ^.     Then 

Pi  P3 

P3  P3 

Now,  let  a  and  b  be  the  points  Zi  and  22  on  the  given  cardioid. 
Then 

P3  P3 

If  we  regard  Si  as  fixed,  22  as  variable,  the  third  vertex  of  the 
triangle,  c,  will  describe   a   cardioid  with   its   cusp   at   the   point 

Pl/e+<p)^^_^^y^  of  size     l+%<^+<^)  !  and  with  orientation- 

P3  P3  I 

angle  [l+^V(«+*')l. 

P3 


40         Cardioids  Fulfilling  Certain  Assigned  Conditions 

For  the  case  of  the  equilateral  triangle,  this  becomes  particularly 
simple.  Then  the  third  vertex,  c,  travels  on  a  cardioid  with  map- 
equation:  c  =  a)(l  —  tiy-\-cjo^{l  —  ty.  This  cardioid  is  of  standard 
size,  with  an  orientation  of  60°  and  its  cusp  at  coil  —  tiY;  i.  e.,  the 
cusp  is  at  the  same  distance  from  0  as  is  Zi,  with  its  vector  turned 
through  an  angle  of  —  60°.  Moreover,  this  c-cardioid  goes  through 
the  point  Zi  on  the  original  cardioid.  Either  of  two  arguments 
may  serve  to  show  this : 

1.  When  the  second  vertex  is  made  to  approach  —  (1  — ^i)-,  the 
third  vertex  does  likewise.  Thus,  the  fixed  vertex,  Zu  towards 
which,  in  the  case  of  an  infinitely  small  triangle,  the  other  two 
vertices  tend,  must  be  on  both  cardioids. 


Fig.  VII. — The  sixth  intersection  is  not  shown. 

2.  Analytically,  if  in  the  equation  above  t  =  tu  then 
c=(o;+co2)(l  — <i)2=  —  (1  — ^i)2;  i.  e.,  zi  is  a  point  on  both  cardioids. 

If  Zi  is  given,  the  number  of  equilateral  triangles  is  the  number 
of  intersections  of  two  cardioids  less  one,  since  the  infinitely  small 
triangle,  all  the  vertices  of  which  are  at  Zi,  should  hardly  be 
regarded  as  a  proper  triangle.  This  number  is  7,  since  the  common 
cusps  at  /  and  J  account  for  8  of  the  16  intersections.  It  happens, 
however,  that  2  of  these  7  intersections  are  always  imaginary;  so 
that  the  maximum  number  of  real  proper  equilateral  triangles 
inscribable  in  a  given  cardioid  is  5.  However,  there  are  not 
always  5;  there  may  be  but  3,  or  only  1. 


('((rdioids  Fulfillinf/  Certain  Assigned  Conditions 


41 


Tliat  there  may  be  as  many  as  5  is  easily  shown:  for,  when  t 
is  a  turn  through  an  angle  numerically  less  than  ±30°,  for  the 
point  05(1-/1)2  the  cusp  of  the  cardioid-locus  of  the  vertex,  c,  is 
inside  the  given  cardioid.  When  ti,  then,  is  a  turn  through  a  very 
small  angle,  there  will  be  positions  where  the  number  of  real 
intersections  will  be  5.  (See  Fig.  VII,  a).  For  one  value  of  ti,  the 
two  curves  will  be  tangent,  and  2  of  the  5  will  coincide  (Fig.  VII,  b) ; 
w  hile  for  most  values  for  which  /i  is  a  turn  through  an  angle  greater 
than  30°  (certainly  for  values  of  6  between  30°  and  330°),  there  is 
})ut  1  real  equilateral  triangle  (Fig.  VII,  c). 


z-fi 


c=ua-t»)**w*(i-t)= 


Fig.  VIII. 

Furthermore,  it  may  be  remarked  that  if  we  now  let  Zi  be  re- 
garded as  variable,  the  cusp  /i-  of  the  cardioid-locus  of  the  third 
vcrl(>x  c  is  given  by  k  =  o)(l  —  tiy~=  —  co[— (1  — /i)'^];  i,  e.,  a  cardioid 
willi  the  same  cusp  and  size  as  the  original,  but  with  an  orientation 
of  —  ()(r.     Hie  relations  of  these  curves  is  shown  in  Fig.  VIII. 

All  hough  wo  have  specialized  for  the  case  of  equilateral  triangles, 
it  is  to  be  observed  that  the  number  of  triangles  of  any  sliape 
inscril)ab](^  in  a  given  cardioid  is  5,  3,  or  1,  according  to  the  position 
of  ,~i;  and  that,  regarding  Zi  as  variable,  the  cusp  of  tlie  r-cardioid 
w  ill  also  describe  a  cardioid. 

We   shall   now   indicate   how  these  considerations   mav   })e  of 


42         Cardioiils  Fid  filling  Certain  A.s.signed  Conditions 

value  in  determining  the  number  of  real  cardioids  on  4  points. 
Let  a,  by  c,  d  be  the  vectors  to  the  given  points.  First,  consider 
the  triangle  formed  by  o,  6,  c.  If  a  be  fixed  on  the  standard  cardi- 
oid  and  b  be  allowed  to  move  along  the  cardioid,  c  will  trace  out  a 
second  cardioid  with  relations  to  the  original  as  indicated  on 
page  39.  Similarly,  fixing  the  point  a  of  the  triangle  formed  by 
rt,  by  dy  on  the  standard  cardioid  and  allowing  b  to  move  along  the 
cardioid,  d.  also,  will  trace  out  a  third  cardioid  with  relations  to 
the  original  as  those  on  page  39,  where  d.  replaces  c,  p4,  ps  rei)lace 
Pi,  p2  and  a,  i3,  5  replace  ip,  yp,  6,  respectively. 

Now  a  mechanism  could  be  devised  by  whicli,  after  having 
placed  the  c-cardioid  and  the  cZ-cardioid  in  relation  to  the  original, 
2i  is  allowed  to  move  along  the  original  curve.  The  cusjjs  of  the 
two  third-vertex-cardioids  will  also  move  on  cardioids,  as  shown 
on  page  39.  As  Zi  moves,  the  second  point,  Z2  (6),  will  not,  in 
general,  be  the  same  point  for  the  c-  and  the  f/-curve;  but  such  a 
coincidence  will  certainly  occur.  When  this  does  happen,  we 
have  a  cardioid  on  the  4  given  points.  x\n  arrangement  could  be 
made  by  Avhich  such  a  coincidence  would  be  indicated;  and  a  simple 
registering  of  these  indications  will  give  the  number  of  real  cardioids 
on  4  points. 


C a rdioids  Fulfilling  Certain  Assigned  Conditions         43 


Vic.  V 


Fig.  IX. — Showing  the  3  real  cardioids  when  the  cusp  and  2  lines 
are  given.  \ 


44         CardioUls  Fulfill  infj  Coin  in  As>iif/Ue(l  Conditions 


Fro.  X  —   lu)N\i:  >;  the  3  vvr.]  Ciir('i(i(l.->  wlien  \hc  ccnlrf  jsiul  2 
lines  are  >  i\en. 


'iG.  XL— Sh()\vin>^-  the  2  real  cardioid-  wIhmi  the  cusp  and  2 
points  are  given. 


VITA 

Sister  Mary  Gervase  Kelley  was  born  in  Roxbury,  Mass.,  Sep- 
tember 8,  1888.  She  received  her  elementary  education  in 
St.  Patrick's  Parochial  School  and  was  graduated  from  the  High 
School  in  1905.  In  1906  she  entered  the  novitiate  of  the  Sisters 
of  Charity,  Halifax,  Nova  Scotia,  and  there  continued  her  studies 
in  the  Novitiate  Normal  School.  From  1908  to  1913  she  taught 
in  the  Halifax  Public  Schools.  In  1910  she  began  work  with  the 
University  of  London,  from  which  institution  she  received  the 
Matriculation  and  the  Intermediate  Arts  certificates.  The  four 
academic  years  since  1913,  with  the  intervening  Summer  Sessions, 
have  been  spent  in  residence  at  the  Catholic  Sisters  College, 
Catholic  University  of  America,  where  she  received  the  degree 
Bachelor  of  Arts  in  1914,  and  that  of  Master  of  Arts  in  1915. 
In  her  graduate  work  the  principal  courses  followed  have  been 
those  under  x\ubrey  E.  Landry,  Ph.D.,  and  John  B.  O'Connor, 
Ph.D.,  for  the  work  done  under  both  of  whom  it  is  the  writer's 
pleasure  to  express  her  appreciation,  and  in  particular  to  acknowl- 
edge gratefully  the  constant  interest  and  kindly  encouragement 
and  assistance  given  by  Dr.  Landry,  not  only  during  the  prepara- 
tion of  this  dissertation,  but  during  her  entire  University  course. 


45 


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